Respuesta :

Answer:

[tex]f^{-1}(x)=\dfrac{1}{3}(\ln x +1)[/tex]

Step-by-step explanation:

Given function:

[tex]f(x)=e^{3x-1}[/tex]

Replace f(x) with y:

[tex]\implies y=e^{3x-1}[/tex]

Take natural logs of both sides:

[tex]\implies \ln y = \ln e^{3x-1}[/tex]

Apply the Power Log Law   [tex]\ln x^n=n\ln x[/tex] :

[tex]\implies \ln y = (3x-1)\ln e[/tex]

As [tex]\ln e=1[/tex] then:

[tex]\implies \ln y = 3x-1[/tex]

Rearrange to make x the subject:

[tex]\implies 3x=\ln y +1[/tex]

[tex]\implies x=\dfrac{1}{3}(\ln y +1)[/tex]

Swap x for [tex]f^{-1}(x)[/tex]  and y for x:

[tex]\implies f^{-1}(x)=\dfrac{1}{3}(\ln x +1)[/tex]

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